document.write( "Question 966336: I need help with this question\r
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Algebra.Com's Answer #590600 by ramkikk66(644) $\"\"$ $\"About$

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Question:
\n" ); document.write( "I am a rectangle my perimeter is 60ft my lenght is twice as long as my width how much area do I cover?\r
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document.write( "Answer:\r\n" );
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document.write( "For a rectangle with width W and length L,\r\n" );
document.write( "Perimeter = 2*(L+W)\r\n" );
document.write( "Area = L*W\r\n" );
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document.write( "Let the width be W ft\r\n" );
document.write( "Then length = 2*W ft\r\n" );
document.write( "Perimeter = 2*(length + width) = 2*(W + 2*W) = 2*3*W = 6*W = 60 ft\r\n" );
document.write( "Solving for W, W = 60/6 = 10 Ft and Length = 2*10 = 20 Ft.\r\n" );
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document.write( "Hence, area = length * width = 20*10 = 200 Sq feet.\r\n" );
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document.write( "Hope this helps!\r\n" );
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